A pass code contains 6 digits. The first 3 digits of the code are all even (2,4,6, or 8) and the last 3 are all odd (1,3,5,7, or 9). If digits can be used more than once, how many possible codes are there?

8000

Step-by-step explanation:

The first 3 digits each have 4 options.

The last 3 digits each have 5 options.

The number of possible codes is:

4 × 4 × 4 × 5 × 5 × 5

= 4³ × 5³

= 8000

A + B = 21

(c) A = −10, B = 31

(d) 24x

2 − 34x + 31

2. (a) -

(b) (x + 3)(2x + 1)(7 − 3x)

(c) y = 1.22

3. (a) 5

(b) f(−2) = 0

(c) (x + 2)(3x + 2)(2x − 1)

4. (a) -

(b) A = −96, B = −48

(c) 3(2x + 1)(x + 4)(x − 4)

5. (a) f(2) = 0

(b) (x − 2)2

(2x + 1)

6. (a) -

(b) (x − 3)(2x − 3)(x